Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 11 - Quadratic Functions and Equations - 11.1 Quadratic Equations - 11.1 Exercise Set - Page 707: 76


.03 (3 percent)

Work Step by Step

We plug in the given values and then solve for $r$ to find: $$ 1000\left(1+r\right)^2=1060.9\\ \left(1+r\right)^2=1.0609\\ r=\sqrt{1.0609}-1\\ r=.03$$
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